a)

Áp dụng định lý Pytago trong $\Delta ABC$ vuông tại $A$

Ta có $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$

$\to B{{C}^{2}}={{6}^{2}}+{{8}^{2}}=100$

$\to BC=10cm$

Ta có ${{S}_{\Delta ABC}}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}AB.AC$

$\to AH.BC=AB.AC$

$\to AH=\dfrac{AB.AC}{BC}=\dfrac{6.8}{10}=4,8cm$

b)

Xét $\Delta AEH$ và $\Delta AHB$, ta có:

+   $\widehat{EAH}$ là góc chung

+   $\widehat{AEH}=\widehat{AHB}=90{}^\circ $

$\to \Delta AEH\backsim\Delta AHB\left( g.g \right)$

c)

Xét $\Delta AFH$ và $\Delta AHC$, ta có:

+   $\widehat{FAH}$ là góc chung

+   $\widehat{AFH}=\widehat{AHC}=90{}^\circ $

$\to \Delta AFH=\Delta AHC\left( g.g \right)$

$\to \dfrac{AF}{AH}=\dfrac{AH}{AC}\to A{{H}^{2}}=AF.AC$

d)

Vì $\Delta AEH\backsim\Delta AHB\left( cmt \right)$

$\to \dfrac{AE}{AH}=\dfrac{AH}{AB}\to A{{H}^{2}}=AE.AB$

Mà $A{{H}^{2}}=AF.AC$

$\to AE.AB=AF.AC$

$\to \dfrac{AB}{AF}=\dfrac{AC}{AE}$

Xét $\Delta ABC$ và $\Delta AFE$, ta có:

+   $\widehat{BAC}$ là góc chung

+   $\dfrac{AB}{AF}=\dfrac{AC}{AE}\left( cmt \right)$

$\to \Delta ABC\backsim\Delta AFE\left( c.g.c \right)$

e)

Ta có ${{S}_{\Delta ABC}}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}\cdot 6\cdot 8=24c{{m}^{2}}$

Với $A{{H}^{2}}=AF.AC\to AF=\dfrac{A{{H}^{2}}}{AC}=\dfrac{4,{{8}^{2}}}{8}=2,88cm$

Vì $\Delta ABC\backsim\Delta AFE\left( cmt \right)$

$\to \dfrac{{{S}_{\Delta ABC}}}{{{S}_{\Delta AFE}}}={{\left( \dfrac{AB}{AF} \right)}^{2}}={{\left( \dfrac{6}{2,88} \right)}^{2}}=\dfrac{625}{144}$

$\to {{S}_{\Delta AFE}}=\dfrac{144}{625}{{S}_{\Delta ABC}}=\dfrac{144}{625}\cdot 24=\dfrac{3456}{625}c{{m}^{2}}$

Vậy ${{S}_{BCFE}}={{S}_{\Delta ABC}}-{{S}_{\Delta AFE}}=48-\dfrac{3456}{625}=\dfrac{26544}{625}c{{m}^{2}}$